Stability of Equilibrium and Buckling Analysis of Struts and Columns
Stability of Equilibrium and Buckling Analysis
Ch.7 notes – Stability of Equilibrium – Consider a slight displacement from the equilibrium position of a rigid weightless bar. The equation -(kLsinΘ)(LcosΘ)-P(LsinΘ)=0 simplifies to P=kLcosΘ or P=kL. Three cases are considered:
- If P < kL, the bar force restores the position.
- If P = kL, the spring force cannot restore the position, and Pc=kL is the critical load.
- When P > Pc, the bar collapses.
Consider an imperfect system: kL(sinΘ-sinΘ0)(LcosΘ)-P(LsinΘ)=0, then P=(kL(sinΘ-sinΘ0)/sinΘ)xcosΘ or P=kL(1-Θ0/Θ). Since sinΘ and cosΘ ≈ 1, Θ=Θ0(1/(1-(P/kL))). Determine the buckling load for the system below.
, Consider the force equilibrium of bar AB.
, P(LΘ)+kLΘ/4 x L/2 -RL=0. For the whole system: (kLΘ)L+kLΘ/4 x L/2+ kLΘ/4 x 3L/2 -2RL=0. P=5kL/8 and 3kLΘ/4. When P>5kL/8, the system is unstable (Buckling of struts/columns). Assume:
- The column is straight.
- The column is loaded through the centroid.
- The column material is linear elastic.
- Buckling or bending occurs in a single plane.
- Analysis is for the critical load for a column in different boundary conditions.
- Approach: treat the column like a horizontal beam rotated at 90 degrees (PIN-ENDED STRUT).
, Consider a slight displaced position, M=Pxu=-EI((d^2)x u)/(dy^2). d^2 x u/dy^2 + P/EI(u)=0 or d^2(u)/dy^2 + k^2(u)=0, where k^2=P/EI. The general equation of u is u=Acosky+Bsin(ky). Boundary conditions: u=0 at y=0 and y=L, then A=0, BsinkL=0. B cannot equal 0, so deflection occurs. Then kL=0, kL=nπ, n=1,2,3. Therefore, P=(n^2 π^2xEI)/L^2. For n=1, Pc=P=π^2(EI)/L^2.
For n=2, Pc=P=4π^2(EI)/L^2. For n=3, Pc=P=9π^2(EI)/L^2. (FIXED-PINNED-STRUT)
Fixed-pinned strut (consider a slightly displaced position), M=P x u – R x y = -EI(d^2 x u/dy^2) or EI(d^2 x u)/dy^2) + P x u = R x y. (d^2 x u/dy^2) + k^2 x u = R/EI(y), where k^2 = P/EI. The general equation is u = Acosky+Bsinky+R/P(y). Boundary conditions: u=0 at y=0 and du/dy=0 at y=L, then A=0, B=-R/Pk x (1/coskL), u=-Rsinky/PkcoskL + R/P(y). With u=0 at y=L, 0=-RsinkL/PkcoskL+R/P(L), tankL=kL. The smallest value of kL for tankL=kL is kL=4.493. k^2=P/EI = 4.493^2/L^2, Pc=20.19EI/L^2 or Pc=π^2(EI)/(0.7L)^2 compared with the pin-ended case. The effective L for the fix-pinned case is Leff=0.7L. (Fix-free strut)
Consider a slightly displaced position, M=-P(δ-u)=-EI(d^2(u))/dy^2. d^2(u)/dy^2 + P/EI = P/EI(δ) or d^2(u)/dy^2 + k^2(u)+k^2(u)=k^2(δ), where k^2=P/EI. The general equation is u=Acosky+Bsinky+δ. Boundary conditions: u=δ at y=0 and y=0 at y=L.
Then A=0, B=-δ/sinkL, du/dy=0 at y=L, then -δ/sinkL(kcoskL)=0. Since δxk cannot equal 0, then 1/tankl=0, kL=nx(π/2) or k^2=n^2(π/2L)^2=P/EI, n=1,3,5. P=n^2xπ^2xEI/(2L)^2 and for n=1, Pc=π^2(EI)/(2L)^2 compared with the pin-ended case. The effective L for the fixed-free case is Leff=2L. (FIX-FIXED Strut) Using the same approach, the following equations can be obtained for the fix-fixed case: Pc=π^2(EI)/(L/2)^2, Leff=L/2. Note that the critical buckling load can be expressed as Pc=π^2(EI)/(Leff)^2, where Leff varies with different end conditions. Use the appropriate I value for buckling reference in different planes. Select the minimum value of I for the cross-section (Iz or Iy, whichever is smaller). For P, use Pc=π^2(EI)/(Leff)^2 if buckling may occur in both the x-y and x-z planes under boundary conditions. (BUCKLING CHARACTER OF REAL STRUTS)
With I=A x r^2, Pc=π^2(EI)/(Leff)^2 = π^2(EA)r^2/(Leff)^2. In terms of stress, σ(c)=Pc/A=π^2(Er^2)/(Leff)^2 = π^2(E)/(Leff/r)^2. The ratio (Leff/r) is the SLENDERNESS RATIO.
Application of Euler theory: In the structure below, tube BC is made of steel with E=207GPa and 340MPa. The inner and outer diameters are 38mm and 48 mm. Bar AB is subjected to a uniformly distributed load of 10kN/m. Determine the factor of safety for buckling.
2Psinα-10x2x1=0, sinα=1/√(1+2^2), P=22.361KN. For Yielding, A=π/4(48^2-38^2)=675mm^2, σ=P/A=22.361×1000/675.1=33.12MPA
For a fix-pinned strut, Pc=π^2EI/(0.7L)^2=20.19EI/L^2. The load at point A is calculated as Pa=200×2=400kN. I=PAL^2/20.19E = 400x100x2000^2/20.19x72x1000=1.101×10^6mm^4. I=π/64(do-di^2)=π/64(75^4-di^2)=1.101×10^6, di=55.1mm, t=do-di/2=75-55.1/2=9.95mm. To avoid buckling, the wall thickness must be >9.95mm. Check stress σ=PA/A=400×1000/(75^2-55.1^2)=196.7MPa. Case A –
M-P(Lθ)=0, M=βR x θ, Pc=βR/L. Case B –
CASE B- RAy=P, Ma=βRxθ, MA=RCx x L, RCx = βRxθ/L
RBx=RCx=βRθ/L , PxθxL/2=RCx x L/2 = βRxθ/2 x L/2, Pc=βR/L END OF CHP.6 Notes
CH.6 NOTES
Moment-Curvature Relation: Consider the deformation of a cantilever beam subject to a concentrated load at the free end. V is the vertical deflection and θ is the rotation/slope of the beam.
ds=Rxdθ. Curvature k=1/R=dθ/ds = dθ/dx. Because dx=ds x cosθ, cosθ=1 when θ is small. δθ=(dv/dx)x+δx – (dv/dx)x. δθ=((dv/dx)x+δx – (dv/dx)x)/δx = d/dx(dv/dx). As δs –> 0, δx–> 0. dθ/dx=d^2(v)/dx^2 = 1/R. Since M/I= E/R= σx/y, then M=-EI(d^2(v))/(dx^2). The negative sign (-) means deflection v is positive when moment M is negative. θ=dv/dx (in rad) will be given by the integration of M=-EI(d^2(v))/(dx^2), which is the rotation/slope of the beam. v will be obtained by the integration of dv/dx, which is the deflection of the beam. (Derivation of Deflection) – Cantilever beam subjected to a point load
M=P(x-L)
M=-EI(d^2 x v)/dx^2, EI(d^2 x v)/dx^2 = -P(x-L), EI(dv/dx) = PLx – 1/2Px^2 + C1, EIv=1/2PLx^2 – 1/6Px^3 + C1x + C2. Apply boundary conditions. v=0 at x=0 and dv/dx = 0 at x=0, then c1=c2 = 0. dv/dx=P/EI(Lx-1/2x^2), v= P/EI(1/2Lx^2 – 1/6x^3).
Maximum deflection v occurs at x=L, vmax= P/EI(1/2L^3 – 1/6L^3) = PL^3/3EI. Maximum slope θ also occurs at x=L. θmax = dv/x| x=L = P/EI(L^2 – 1/2L^2)=PL^2/2EI.
(CANTILEVER Beam subjected to a uniformly distributed load)
EI(dv/dx) = PLx – 1/2Px^2 + C1, EIv=1/2PLx^2 – 1/6Px^3 + C1x + C2. Apply boundary conditions. v=0 at x=0 and dv/dx = 0 at x=0, then c1=c2 = 0. dv/dx=P/EI(Lx-1/2x^2), v= P/EI(1/2Lx^2 – 1/6x^3).
Maximum deflection v occurs at x=L, vmax= P/EI(1/2L^3 – 1/6L^3) = PL^3/3EI. Maximum slope θ also occurs at x=L. θmax = dv/x| x=L = P/EI(L^2 – 1/2L^2)=PL^2/2EI.
(CANTILEVER Beam subjected to a uniformly distributed load)
RA=wL, M=RAx-MR – wx^2/2=wLx-wL^2/2-wx^2/2. EI(d^2(v))/dx^2 = -wLx+wL^2 + wx^2/2. EI(dv/dx)=-wL/2x^2 + wL^2/2(x) + wx^3/6 +C1. EIv=-wL/69(x^3)+wL^2/4(x^2)+wx^4/24+C1x+C2. Apply boundary conditions – v=0 at x=0 and dv/dx at x=0, then C1=C2=0. v=w/EI(-L/6(x^3)+L^2/4(x^2)+1/24(x^4)), θ=w/EI(-L/2(x^2)+L^2/2(x)+1/6(x^3)). Maximum deflection v occurs at x=L, Vmax= wL^4/8EI, Maximum slope occurs also occurs at x=L, θmax=wL^3/6EI.
(SIMPLY supported beam under uniformly-distributed load)
M=0.5w(L x (X-X^2)), EI(d^2(v))/dx^2)=1/2wx^2-1/2wLx, EI(dv/dx)=1/6wx^3-1/4wLx^2 +C1, EI(v)=1/24wx^4-1/12wLx^3+C1(x)+C2, Apply BCS – v=0 at x=0 and x=l, and dv/dx=0 at x=L/2, then C1=wL^3 and C2=0. Thus v=-w/12EI(Lx^3 -x^4/2 – L^3(x)/2) maximum deflection occurs at x=L/2, vmax=-w/12EI(L(L/2)^3-1/2(L/2)^4 – L^3/2(L/2)) = 5/384 x wL^4/EI, θ=w/2EI(x^3/3 – Lx^2/2+L^3/12), Maximum slope occurs at x= 0 (positive) and x=L (negative). θMax = dv/dx | x=0 = wL^3/24(EI) and θmin=dv/dx| x=L = -wL^3/24EI.
(Simply supported beam under concentrated load)
When x